2 years ago

Find the exterior angle

1 answer:

wel2 years ago

6 0

Check the picture below.

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Bond [772]

$f(x)=3\sin x+3\cos x\implies f'(x)=3\cos x-3\sin x$

$f$ has critical points where $f'=0$ :

$3\cos x-3\sin x=0\implies\cos x=\sin x\implies\tan x=1\implies x=\pm\dfrac\pi4+2n\pi$

where $n$ is any integer. We get solutions in the interval $\left[0,\frac\pi3\right]$ for $n=0$ , for which $x=\frac\pi4$ .

At this critical point, we have $f\left(\frac\pi4\right)=3\sqrt2\approx4.243$ .

At the endpoints of the given interval, we have $f(0)=3$ and $f\left(\frac\pi3\right)=\frac{3+3\sqrt3}2\approx4.098$ .

So we have the extreme values

$\max\limits_{x\in\left[0,\frac\pi3\right]}f=3\sqrt2$

$\min\limits_{x\in\left[0,\frac\pi3\right]}f=3$

8 0

3 years ago

Damm [24]

Answer:

-3 is the answer for the question

5 0

2 years ago

solong [7]

Answer:

138.80

Step-by-step explanation:

I took 149.99 times 15% then times that by the tax rate to get your final answer.

8 0

2 years ago

Aleks [24]

F(x) = x^3 + 2x^2 - x - 2
f(x) = x^2(x + 2) - (x + 2) 
f(x) = (x^2 - 1)(x + 2) 
f(x) = (x + 1)(x - 1)(x + 2) 
Roots: -1, 1, -2

3 0

3 years ago

yulyashka [42]

Answer: Sorry

Step-by-step explanation: What table? I would love to help you but I need to see the table you are talking about. :(

5 0

2 years ago

Find the exterior angle​